Exercise 2.72

Answers

(a)

p_{n} = a_{n} a + (1 - a_{n}) b = (a - b) a_{n} + b

a_{n + 1} = a_{n} a + (1 - a_{n}) (1 - b) = a_{n} (a + b - 1) + 1 - b

(b)

p_{n + 1} = (a - b) a_{n + 1} + b

p_{n + 1} = (a - b) ((a + b - 1) a_{n} + 1 - b) + b

p_{n + 1} = (a - b) ((a + b - 1) \frac{p_{n} - b}{a - b} + 1 - b) + b

p_{n + 1} = (a + b - 1) p_{n} + a + b - 2 ab

(c)

Let

p = \lim_{n \to \infty} p_{n}

. Taking the limit of both sides of the result of part

b

, we get

p = (a + b - 1) p + a + b - 2 ab

p = \frac{a + b - 2 ab}{2 - (a + b)}

fifthist

2021-12-05 00:00